A ray of light suffers minimum deviation, while passing through a prism of refractive index `1.5` and refracting angle `60^@`. Calculate the angle of deviation and angle in incidence.

Answer

`37.2^(@), 48.6^(@)`

Explanation

`mu = 1.5, A = 60^(@), delta_(m) = ?, i = ?`
As `(sin (A + delta_(m))//2)/(sin A//2) = mu`
`:. (sin (60^(@) + delta_(m))//2)/(sin 60^(@)//2) = 1.5`
`sin (60^(@) + delta_(m))//2 = 1.5 sin 30^(@) = 0.75`
`(60^(@) + delta_(m))/(2) = sin^(-1)(0.75) = 48.6^(@)`
`delta_(m) = 2 xx 48.6 - 60^(@) = 37.2^(@)`
`i = (A + delta_(m))/(2) = (60^(@) + 37.2)/(2) = 48.6^(@)`